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What is the antiderivative of $(2x+7)^{1/2}$?

My understanding is that it would be $\frac 23 \times(2x+7)^{3/2}$ but according to the source I am working from the answer is $\frac 13 \times (2x+7)^{3/2}$.

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    $\begingroup$ Differentiate your answer and you will see why. HINT: Chain rule. $\endgroup$
    – mickep
    Commented Mar 2, 2016 at 10:00
  • $\begingroup$ You probably forgot to process the coefficient of $x$. $\endgroup$
    – user65203
    Commented Mar 2, 2016 at 10:13
  • $\begingroup$ Why do I need to process the coefficient of x before I take the antiderivative? $\endgroup$
    – user252704
    Commented Mar 2, 2016 at 11:57
  • $\begingroup$ Because it is going to be processed when you take the derivative. If it changes in one direction, it has to change in both directions, or else the derivative of an antiderivative wouldn't equal the original function. $\endgroup$
    – kevinsa5
    Commented Mar 2, 2016 at 14:13

3 Answers 3

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$$\int\sqrt{(2x+7)}dx$$

Set $t=2x+7$ and $dt=2dx$

$$=\frac 1 2\int\sqrt tdt=\frac{t^{3/2}}{3}+\mathcal C=\frac{(2x+7)^{3/2}}{3}+\mathcal C$$

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  • $\begingroup$ Why multiply it by 1/2 and also by 2? $\endgroup$
    – user252704
    Commented Mar 2, 2016 at 12:01
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    $\begingroup$ @user252704 its because $dt=2dx$...therefore $dx=\dfrac{dt}{2}$...now all we need to do is to plug in the values... $\endgroup$
    – manshu
    Commented Mar 2, 2016 at 12:39
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Notice:

$$\int x^n\space\text{d}x=\frac{x^{1+n}}{1+n}+\text{C}$$



$$\int\sqrt{2x+7}\space\text{d}x=$$


Substitute $u=2x+7$ and $\text{d}u=2\space\text{d}x$:


$$\frac{1}{2}\int\sqrt{u}\space\text{d}u=\frac{1}{2}\int u^{\frac{1}{2}}\space\text{d}u=\frac{1}{2}\cdot\frac{u^{1+\frac{1}{2}}}{1+\frac{1}{2}}+\text{C}=\frac{u^{\frac{3}{2}}}{3}+\text{C}=\frac{(2x+7)^{\frac{3}{2}}}{3}+\text{C}$$

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  • $\begingroup$ Why multiply it by 1/2 and also by 2? $\endgroup$
    – user252704
    Commented Mar 2, 2016 at 12:01
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    $\begingroup$ To make the substitution clear $\endgroup$ Commented Mar 2, 2016 at 12:59
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$$(2x+7)^{1/2}=2^{1/2}(x+\frac72)^{1/2}\to2^{1/2}\frac1{\frac32}(x+\frac72)^{3/2}=\frac{2^{3/2}}3(x+\frac72)^{3/2}=\frac{(2x+7)^{3/2}}3.$$

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  • $\begingroup$ Why do I need to process the coefficient of x before I take the antiderivative? $\endgroup$
    – user252704
    Commented Mar 2, 2016 at 11:58
  • $\begingroup$ I noticed that you didn't handle it properly so this is a workaround. $\endgroup$
    – user65203
    Commented Mar 2, 2016 at 12:13
  • $\begingroup$ @user252704, it's like the chain rule in reverse... (2x+7)^(1/2) is like f(g), where f(x) = x^(1/2) and g(x) = 2x+7, except here you have to remember to process the ANTIderivative of g instead of the derivative $\endgroup$
    – sig_seg_v
    Commented Mar 2, 2016 at 16:50

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